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A134835
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Let {b_n(m)} be a sequence defined by b_n(0)=0, b_n(m) is the largest prime dividing (b_n(m-1) + n). Then a(n) is the smallest positive integer such that b_n(m + a(n)) = b_n(m), for all integers m that are greater than some positive integer M.
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1
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1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 5, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 22, 1, 5, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 6, 1, 1, 1, 6, 1, 1, 1, 1, 1, 10, 1, 1, 1, 8, 1, 1, 1, 1, 1, 7, 1, 9, 1, 1, 1, 1, 1, 1, 1, 5, 14, 1, 1, 6, 1, 1, 1
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OFFSET
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2,3
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LINKS
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EXAMPLE
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Sequence {b_8(m)} is 0, 2, 5, 13, 7, 5, 13, 7, ... (5, 13, 7) repeats. So a(8) = 3, the length of the cycle in {b_8(m)}.
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PROG
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(PARI) a(n) = my(b, k, v=List([0])); until(k<#v, k=1; listput(v, b=vecmax(factor(b+n)[, 1])); until(v[k]==b||k==#v, k++)); #v-k; \\ Jinyuan Wang, Aug 22 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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